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The degree-one Bockstein homomorphism makes modulo-$n$ cohomology into a cochain complex. Its cohomology is the Bockstein cohomology. For prime $n$ this is a graded vector space; for composite $n$ it is a graded $\mathbb{Z}/n$-module. For ${\mathbb{RP}}^3$ at modulus two, its only nonzero groups are ${\mathbb{F}}_2$ in degrees zero and three. This is the cohomological counterpart of Bockstein homology.

For the modulo-$m$ cohomological Bockstein homomorphism,This follows by lifting cocycles to integral cochains and applying the graded Leibniz rule for the coboundary before dividing by $m$.

The Bockstein homomorphism for $0\to \mathbb{Z}/n\to \mathbb{Z}/n^2\to \mathbb{Z}/n\to 0$, with injection $[a]\mapsto [na]$, is coefficient reduction $\rho$ after the integral Bockstein homomorphism. The commuting maps of coefficient sequences prove this by naturality of the connecting homomorphism.

Attach an $(i+1)$-cell to $S^i$ by a degree-$n$ map, with $i\ge 1$ and $n\ge 2$. The integral cellular differential is multiplication by $n$, while its modulo-$n$ reduction is zero. Lifting the degree-$i$ generator modulo $n^2$ and applying the differential gives $n$, so the Bockstein homomorphism sends that generator to the degree-$(i+1)$ generator. This supplies a nonzero example in every positive degree.

For the coefficient sequence $0\to {\mathbb{Z}}_2\to {\mathbb{Z}}_4\to {\mathbb{Z}}_2\to 0$, the Bockstein homomorphism on infinite-dimensional real projective space is zero from an even degree and an isomorphism from an odd degree. In cellular cohomology, lift the mod-two generator to $1\in {\mathbb{Z}}_4$. The coboundary is zero in even degree and two in odd degree; identifying two with the image of $1\in {\mathbb{Z}}_2$ gives the formula.

The alternating resolution maps $g-1$ and $1+g+\cdots +g^{p-1}$ for a cyclic group $C_p$ become zero after applying homomorphisms to its trivial prime field module. Thus there is one class in every degree. A two-step shift of the resolution represents a degree-two periodicity class $u$. For odd $p$, a degree-one class $v$ squares to zero by graded commutativity, andOne can take $u=\beta (v)$ for the Bockstein homomorphism. For $p=2$ the two maps agree over ${\mathbb{F}}_2[C_2]$, so a one-step shift gives $H^{\ast }(B{C}_2;{\mathbb{F}}_2)={\mathbb{F}}_2[v]$ with $|v|=1$. The Künneth theorem then gives dimension $i+1$ in degree $i$ for $B(C_p\times C_p)$, obstructing periodic group cohomology and hence such a free sphere action.

A long exact sequence is a sequence of abelian groups or modules with arbitrarily many consecutive maps, for which the image of every map equals the kernel of the next. A short exact sequence of cochain complexes yields a long exact sequence of their cohomology, with connecting homomorphisms increasing degree by one. The exact sequences of relative cohomology and the Bockstein homomorphism are examples.

A short exact sequence of coefficient groups $0\to A\to B\to C\to 0$ induces a natural long exact sequenceThe connecting map is a Bockstein homomorphism.

Use the short exact sequence of coefficient groupsThe groups of singular chains are free, so applying the cochain construction gives a short exact sequence of cochain complexes. Its long exact sequence in cohomology has the required terms and reduction map $r_{\ast }$. Define the Bockstein homomorphism to be its connecting homomorphism.

Explicitly, represent $x$ by a mod-$p$ cocycle $a$ and lift it to an integral singular cochain $A$. Since $a$ is a cocycle, $\delta A$ is divisible by $p$. ThenThis is exactly the lift-and-coboundary construction for the coefficient sequence above: the mod-$p^2$ coboundary of $A$ lies in the image of $j$. The quotient cochain is a cocycle, since ${\delta }^{2}A=0$ and integral cochains have no $p$-torsion. Replacing $A$ by another lift adds $pC$ and changes the quotient by $\delta C$ modulo $p$; replacing the representative by a coboundary likewise leaves its cohomology class unchanged. Thus the definition is independent of choices.

A degree-zero cocycle is a function on points constant on each path component. Choose an integer representative for its value on each component. This produces an integral zero-cocycle lift, with zero coboundary. Hence the Bockstein vanishes on $H^0(X;{\mathbb{F}}_p)$ for every space.

For a nonzero example in every positive degree $i$, take the Moore space $X=S^i{\cup }_p{e}^{i+1}$, attaching the cell by a map of mapping degree $p$. Its positive-degree integral cellular cochain complex has the differential $\mathbb{Z}\stackrel{p}{\to }\mathbb{Z}$ from degree $i$ to $i+1$; when $i=1$, the preceding differential from degree zero is zero. Modulo $p$ the displayed differential vanishes, and $H^i(X;{\mathbb{F}}_p)=H^{i+1}(X;{\mathbb{F}}_p)={\mathbb{F}}_p$. Lift the degree-$i$ generator to the cellular cochain with value one. Its coboundary has value $p$, so dividing by $p$ gives the generator in degree $i+1$. Naturality of cellular and singular cohomology with coefficient sequences identifies this with the Bockstein on a cyclic Moore space. Therefore $\beta$ is nonzero, indeed an isomorphism, in the requested degree for every prime $p$.

Finally take integral lifts $A,B$ of cocycles representing $x\in H^i(X;{\mathbb{F}}_p)$ and $y\in H^j(X;{\mathbb{F}}_p)$. Write $\delta A=p{A}_1$ and $\delta B=p{B}_1$. The cochain cup product obeys the graded Leibniz identity, soDivide by $p$, reduce modulo $p$ and pass to cohomology. This proves the Bockstein derivation rule

Unless a coefficient group is displayed, use integral singular cohomology. The standard CW complex structure on infinite-dimensional real projective space has one cell in each nonnegative dimension. Its cellular chain complex has boundary $d_j=2$ for positive even $j$ and $d_j=0$ for odd $j$. The cellular cohomology differential ${\delta }^j$ is therefore zero for even $j$ and multiplication by two for odd $j$. ConsequentlyMore generally, for an abelian group $A$, the positive odd groups are $A[2]=\{a:2a=0\}$ and the positive even groups are $A/2A$. In particular $H^j({\mathbb{RP}}^{\infty };{\mathbb{Z}}_2)={\mathbb{Z}}_2$ in every nonnegative degree.

For the required Bockstein homomorphism, use the short exact sequenceHere $m\ge 2$; for $m=1$ all three coefficient groups are zero. Since singular chains are free abelian groups, applying cochains gives a short exact sequence of cochain complexes. Its connecting homomorphism defines $\beta$ and its long exact sequence is precisely the required one, with the other maps induced by $\iota$ and $q$.

Explicitly, represent a class by a cocycle $u\in C^n(X;{\mathbb{Z}}_m)$ and choose a lift $\tilde{u}\in C^n(X;{\mathbb{Z}}_{m^2})$. Since $q(\delta \tilde{u})=0$, there is a unique cochain $v$ with $\iota (v)=\delta \tilde{u}$. Injectivity of $\iota$ and ${\delta }^2=0$ show $\delta v=0$. DefineChanging the lift by $\iota (w)$ changes $v$ by the coboundary $\delta w$. Changing the representative $u$ by a coboundary can be lifted by a coboundary as well and leaves the resulting class unchanged. Thus this is a well-defined group homomorphism, and the standard cochain lifting argument gives exactness.

Compute the Bockstein homomorphism on infinite-dimensional real projective space using its cellular cohomology complex. A generator with coefficients ${\mathbb{Z}}_2$ is represented by $1$ in degree $n$, lifted to $1\in {\mathbb{Z}}_4$. Its coboundary is $0$ for even $n$ and $2$ for odd $n$. Dividing via $\iota (1)=2$ givesThus the odd-degree maps are isomorphisms. The comparison between cellular cohomology and singular cohomology is natural with respect to coefficient maps, so this computes the same connecting homomorphism constructed above.

Write $u=u_n\in H^n(K(\mathbb{Z}/2,n);{\mathbb{F}}_2)$ for the fundamental class. We use the path-loop Serre spectral sequence, its multiplicative structure, and the Kudo transgression theorem: transgression of the fundamental classes and their compatible Steenrod squares is natural, equivalently cohomology suspension commutes with these stable operations. The initial ring is $H^{\ast }(K(\mathbb{Z}/2,1);{\mathbb{F}}_2)={\mathbb{F}}_2[t_1]$.

Here is the low-degree calculation. In the first path fibration, $t$, $t^2$ and $t^4$ transgress respectively to $u_2$, ${\mathrm{Sq}}^1{u}_2$ and ${\mathrm{Sq}}^2{\mathrm{Sq}}^1{u}_2$, in degrees $2,3,5$. These supply all indecomposable base classes through degree five; the other classes there are products. Repeating the path-loop calculation raises the degree of the transgressive classes by one. For $n=3$, the class ${\mathrm{Sq}}^2{u}_3$ transgresses from $u_2^2$, and ${\mathrm{Sq}}^2{\mathrm{Sq}}^1{u}_3$ from the degree-five generator. For $n=4$, ${\mathrm{Sq}}^3{u}_4$ transgresses from $u_3^2$. For larger $n$ all four displayed low-degree operations are transgressive indecomposables. Acyclicity of the path-space total cohomology forces these transgressions and excludes additional classes in this range.

More systematically, this is the range up to $n+3$ of the Serre polynomial generators for mod-two Eilenberg–MacLane cohomology:For an admissible sequence of Steenrod squares, $I=(i_1,\dots ,i_r)$ has $i_j\ge 2i_{j+1}$; its degree increment is $\sum i_j$, and its Steenrod excess is $e(I)=i_1-i_2-\cdots -i_r$. Include the empty sequence. Up to increment three the nonempty possibilities are $(1),(2),(3),(2,1)$; the strict excess bound and possible products explain precisely the small-$n$ exceptions.

The low-degree mod-two cohomology of an Eilenberg–MacLane space isNegative-degree cohomology is zero. For $n=2$, ${\mathrm{Sq}}^{2}u=u^2$ and ${\mathrm{Sq}}^{3}u=0$ by instability, so the product in degree five must not be omitted. For $n=3$, ${\mathrm{Sq}}^{3}u=u^2$ is a product, still independent from ${\mathrm{Sq}}^2{\mathrm{Sq}}^{1}u$. For $n\ge 4$ the listed classes are indecomposable. The Adem relations include ${\mathrm{Sq}}^1{\mathrm{Sq}}^1=0$ and ${\mathrm{Sq}}^1{\mathrm{Sq}}^2={\mathrm{Sq}}^3$, so no further increment-three class comes from reversing the two squares.

Now use the space actually printed in the PDF,the converted TeX dropped the projective-space $P$'s. This stunted real projective space has one cell in dimensions $7,8,9,10$, besides its basepoint, and is $6$-connected. The integral cellular boundary is $2$ in even dimensions and zero in odd dimensions. HenceThe Hurewicz theorem gives ${\pi }_7(Y)=\mathbb{Z}/2$.

For the next two groups, let $K=K(\mathbb{Z}/2,7)$ and choose $f:Y\to K$ representing the generator of $H^7(Y;{\mathbb{F}}_2)$. It induces an isomorphism on ${\pi }_7$, and the target's higher homotopy groups vanish.

We need the low-degree integral homology of a mod-two Eilenberg–MacLane space. Here it isTo justify the torsion orders, the Serre class theorem first makes every positive-degree integral homology group of $K$ a finite 2-group. The universal coefficient theorem for cohomology with ${\mathbb{F}}_2$ and the low-degree dimensions give one cyclic summand in degrees seven, nine and ten and none in degree eight. The mod-two Bockstein homomorphism is ${\mathrm{Sq}}^1$. Its relevant nonzero actions areThe last target is nonzero: $(3,1)$ is admissible with excess two, below seven. It also follows by transgressing the nonzero square $({\mathrm{Sq}}^1{u}_2{)}^2$ through successive path fibrations. Each nonzero Bockstein pairs the mod-two classes associated with a cyclic integral summand of order exactly two; for a summand of order $2^r$ with $r>1$, this first Bockstein would be zero. In degree ten, ${\mathrm{Sq}}^{3}u$ is already the Ext class from $H_9$, so the independent class ${\mathrm{Sq}}^2{\mathrm{Sq}}^{1}u$ detects $H_{10}$. This proves all four integral groups without confusing them with mod-two Betti numbers.

Let $x\in H^1({\mathbb{RP}}^{10};{\mathbb{F}}_2)$ be its usual generator. The quotient classes in degrees $7$ through $10$ identify with $x^7,\dots ,x^{10}$ via the pair's cohomology. The Steenrod squares on real projective space satisfyThussince $\left(\genfrac{}{}{0ex}{}{7}{2}\right)$ and $\left(\genfrac{}{}{0ex}{}{7}{3}\right)$ are odd, while $\left(\genfrac{}{}{0ex}{}{8}{2}\right)$ is even. In degree nine, the nonzero cohomology map detects the map on $H_9$, as $H_8=0$ on both sides. Consequently $f_{\ast }:H_9(Y;\mathbb{Z})\to H_9(K;\mathbb{Z})$ is an isomorphism.

Treat $f$ as a mapping-cylinder pair $(K,Y)$. It is $8$-connected: both spaces are $6$-connected, ${\pi }_7$ is an isomorphism, and ${\pi }_8(K)=0$. The relative homology sequence gives $H_9(K,Y)=0$, since $H_9(Y)\to H_9(K)$ is an isomorphism and $H_8(Y)=0$. The Relative Hurewicz theorem then gives ${\pi }_9(K,Y)=0$, and the relative homotopy sequence identifies this group with ${\pi }_8(Y)$. Thus ${\pi }_8(Y)=0$ and the pair is now $9$-connected.

Next,because $H_{10}(Y)=0$ and the degree-nine map is an isomorphism. Apply the Relative Hurewicz theorem again and use ${\pi }_{10}(K)={\pi }_9(K)=0$:Therefore the homotopy groups of the stunted projective space through degree nine are

First account for the unlabelled coefficient-sequence construction. The two short exact sequences of abelian groups areThe singular chain groups of $X$ are free abelian. Applying $\mathrm{Hom}(C_q(X),-)$ therefore preserves these exact sequences, degree by degree, giving short exact sequences of cochain complexes. The associated long exact sequence from a coefficient sequence gives the displayed maps in cohomology; the connecting maps are the integral Bockstein homomorphism $\hat{\beta }$ and the modulo-$n$ Bockstein homomorphism $\beta$. The first omitted map is multiplication by $n$, and the second is induced by $j$.

For the requested example, attach an $(i+1)$-cell to $S^i$ using a map of degree $n$. The resulting Moore space $X=M(\mathbb{Z}/n,i)$ has positive-degree cellular chain complexin degrees $i+1,i$. This construction also works for $i=1$, using the degree-$n$ map of the circle. In cellular cohomology with coefficients $\mathbb{Z}/n$, the differential is zero, so both $H^i$ and $H^{i+1}$ are $\mathbb{Z}/n$.

Lift the cochain taking value $1$ on the $i$-cell to a cochain with coefficients $\mathbb{Z}/n^2$. Its coboundary takes value $n$ on the $(i+1)$-cell, which is $j(1)$. The definition of the connecting homomorphism therefore sends the degree-$i$ generator to the degree-$(i+1)$ generator. HenceIt is nonzero for every $i\ge 1$ and $n\ge 2$, including composite $n$. This is the Bockstein on a cyclic Moore space.

The standard CW complex structure on Real projective space has one cell in each dimension from zero to three. Its integral cellular boundary is multiplication by $2$ in even positive degrees and zero in odd degrees. Thus the integral cellular cochain complex for ${\mathbb{RP}}^3$ isin degrees $0,1,2,3$. With coefficients ${\mathbb{F}}_2$, all its differentials vanish, so every one of these four cohomology groups is one-dimensional.

The lift-and-divide construction of the Bockstein homomorphism turns the integral differential $2$ into $1$ modulo $2$. Therefore $\beta :H^1\to H^2$ is an isomorphism, while the maps from degrees $0,2,3$ are zero. The Bockstein cohomology is consequentlyFor comparison, in the mod-two cohomology ring of real projective space ${\mathbb{F}}_2[t]/(t^4)$, $|t|=1$, this says $\beta (t)=t^2$, $\beta (t^2)=0$ and $\beta (t^3)=0$. The last two formulas also follow from the Bockstein derivation rule and the truncation $t^4=0$.

Choose a contractible free $G$-space $EG$, with $BG=EG/G=K(G,1)$ the classifying space of a discrete group. The Borel construction gives the fibrationBecause the action on the sphere is free, the map $E\to S^{n-1}/G$ has contractible fibre $EG$ and is a homotopy equivalence. The quotient is a closed $(n-1)$-dimensional manifold, so $H^k(E;\mathbb{F})=0$ for $k>n-1$.

The hypothesis on integral sphere cohomology makes the $G$-action on $H^{n-1}(S^{n-1};\mathbb{F})$ trivial as well. Thus the Serre spectral sequence has just two rows, both copies of $H^{\ast }(BG;\mathbb{F})$, in fibre degrees zero and $n-1$. If $u$ is the fibre generator, the only possible differential is $d_n$, and its transgression definesThe multiplicative differential rule gives $d_n(au)=(-1{)}^{|a|}a⌣\Delta$. Up to the graded sign, this is multiplication by $\Delta$. For $i>0$, the terms $E_{\infty }^{i,n-1}$ and $E_{\infty }^{i+n,0}$ must both vanish because their total degrees exceed $n-1$. The first vanishing says the relevant multiplication map has zero kernel; the second says it has zero cokernel. ThereforeThis is periodic group cohomology from a free sphere action. Degree zero was excluded for a reason: $E_{\infty }^{0,n-1}$ can contribute to the quotient's top cohomology.

For $C_p=\mathbb{Z}/p$, use its periodic free resolution with alternating maps $g-1$ and $N=1+g+\cdots +g^{p-1}$. Applying ${\mathrm{Hom}}_{\mathbb{Z}[C_p]}(-,{\mathbb{F}}_p)$ makes every differential zero, so $H^i(B{C}_p;{\mathbb{F}}_p)$ is one-dimensional in every nonnegative degree. The two-step shift of the resolution supplies a nonzero degree-two class $u$ whose cup product gives the periodicity isomorphisms. For odd $p$, let $v$ be a nonzero degree-one class. Graded commutativity gives $v^2=0$, and the shift gives nonzero $v{u}^k$ and $u^k$ in every degree. For $p=2$, the two resolution maps agree over ${\mathbb{F}}_2[C_2]$, giving a one-step shift whose degree-one class has nonzero powers. Hence the cohomology ring of a finite cyclic group over its prime field isHere the exterior factor is an exterior algebra and the polynomial factor is a polynomial ring. For odd $p$, $u$ may be chosen as the Bockstein homomorphism of $v$.

If $G$ contained $C_p\times C_p$, restriction would give that subgroup the same free, cohomologically trivial sphere action. It would therefore also have period $n$ in positive-degree cohomology. But $B(C_p\times C_p)\simeq B{C}_p\times B{C}_p$, and the Künneth theorem givesThese dimensions strictly increase, so the groups in degrees $i$ and $i+n$ cannot be isomorphic. Thus $G$ contains no subgroup isomorphic to $\mathbb{Z}/p\times \mathbb{Z}/p$ for any prime $p$.

Put $F={\mathbb{F}}_2$. Under the usual bundle convention of a paracompact base, choose a fiber metric and let $D(E),S(E)$ be the disk bundle and sphere bundle of the rank-$n$ real vector bundle. A mod-two Thom class is a relative cohomology classwhose restriction to every fiber pair $(D^n,S^{n-1})$ is the nonzero generator. Equivalently it is a class in $H^n(E,E\setminus B;F)$, where $B$ is identified with the zero section. No orientation choices are needed over $F$.

The Thom isomorphism theorem asserts that this class exists and that, for every $r$, the mapis an isomorphism. The disk bundle retracts to the zero section $s:B\to D(E)$, so $H^r(D(E);F)\cong H^r(B;F)$. Define the mod-two Euler class $e_2(E)=s^{\ast }u\in H^n(B;F)$, where the pullback includes the passage from relative to absolute cohomology. Under the Thom identification, the map from relative to absolute cohomology sends $a$ to $a⌣e_2(E)$. Substituting into the pair's long exact sequence in cohomology gives the Gysin sequence of a sphere bundle:Here ${\pi }_{!}$ is the connecting homomorphism followed by ${\Phi }^{-1}$. Thus the sequence and the cup-product map have been derived from the Thom theorem, not just stated. The Thom/Gysin constructions are also discussed in Hatcher's Vector Bundles and K-Theory, §3.2.

The original PDF specifies the tautological bundle on ${\mathbb{RP}}^n$; the occurrences of ${\mathbb{R}}^n$ in this part of the TeX transcription have lost the projective-space symbol. Let $L\to {\mathbb{RP}}^n$ be this real line bundle. Its sphere bundle consists of pairs $(\ell ,v)$ with $v$ a unit vector in the line $\ell$, so $S(L)\cong S^n$, with projection the antipodal double covering. Write $t=e_2(L)$.

The additive mod-two groups, obtainable from the one-cell-in-each-dimension cellular chain complex, areFor $n\ge 1$, both the base and $S^n$ are connected, so ${\pi }^{\ast }:H^0({\mathbb{RP}}^n;F)\to H^0(S^n;F)$ is an isomorphism. Exactness makes the following ${\pi }_{!}$ zero and multiplication by $t$ injective from $H^0$ to $H^1$. For $1\le r<n$, the term $H^r(S^n;F)$ vanishes, so the same Gysin sequence makesinjective. Since these groups are one-dimensional, the maps are isomorphisms. Hence $1,t,\dots ,t^n$ are the respective nonzero generators, while $t^{n+1}=0$ by dimension. The mod-two cohomology ring of real projective space is thereforeFor $n=0$ this is simply $F$, with $t=0$.

For integral coefficients, the additive groups areIndeed, the cellular homology of real projective space has boundary multiplication by $2$ in even positive chain degrees and zero in odd degrees, and the integral cochain differential is its transpose. It remains to determine the products; the additive groups alone do not do so.

For $n\ge 2$, the coefficient sequence $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to F\to 0$ has a Bockstein homomorphism $\beta$. Because $H^1({\mathbb{RP}}^n;\mathbb{Z})=0$ and $H^2({\mathbb{RP}}^n;\mathbb{Z})=\mathbb{Z}/2$, exactness makes $a=\beta (t)$ the nonzero integral degree-two class. Reduction $\rho :H^2({\mathbb{RP}}^n;\mathbb{Z})\to H^2({\mathbb{RP}}^n;F)$ is injective: its kernel is the image of multiplication by $2$, which is zero. ThusMore generally, reduction is injective in each positive even degree, and its compatibility with the cup product gives $\rho (a^j)=t^{2j}$. Consequently $a^j$ is the nonzero generator whenever $2j\le n$. Also $2a=0$, and powers beyond the dimension vanish.

If $n=2m$ is even, these powers and the unit account for all groups, so the integral cohomology ring of real projective space isIf $n=2m+1$ is odd, add an integral top-dimensional orientation class $b$ of degree $2m+1$. Its reduction is the nonzero class $t^{2m+1}$. Every product $ab$ and $b^2$ is zero by dimension, and $b$ has infinite additive order. HenceThese formulas include $m=0$: the even space is a point with ring $\mathbb{Z}$, and the odd space is a circle with ring $\mathbb{Z}[b]/(b^2)$, $|b|=1$. The polynomial presentations are interpreted with the displayed grading and products; no extra positive-degree products are left unspecified.

An $R$-orientation of a rank-$d$ real vector bundle $E\to B$ is a locally coherent choice of generator of $H^d(E_b,E_b\setminus \{0\};R)$ in every fibre. Equivalently, it is represented by a Thom class $u_E\in H^d(D(E),S(E);R)$ restricting to the chosen generator on each fibre pair. If $e(E)\in H^d(B;R)$ is the Euler class, the Gysin sequence of its unit sphere bundle contains

The diagonal quotient defining $L(p)$ is the three-dimensional lens space as a circle bundle $S^1\to L(p)\to S^2$ with Euler class $p$ times a generator. With integral coefficients, the only nontrivial Euler-class map is multiplication by $p:H^0(S^2;\mathbb{Z})\to H^2(S^2;\mathbb{Z})$. Exactness givesModulo $p$, the Euler class vanishes, so the same Gysin sequence giveswith all other groups zero.

For the coefficient sequence $0\to \mathbb{Z}\stackrel{p}{\to }\mathbb{Z}\to {\mathbb{F}}_p\to 0$, the long exact sequence from a coefficient sequence containsThe last map is zero and both middle groups have order $p$, so $\delta$ is an isomorphism. Reduction $H^2(L(p);\mathbb{Z})\to H^2(L(p);{\mathbb{F}}_p)$ is likewise an isomorphism. Their composite is the Bockstein isomorphism for a three-dimensional lens space

If $a^{\prime }=na$ is another generator, linearity of the Bockstein homomorphism and bilinearity of the cup product giveMoreover $t(a)\ne 0$: $\beta (a)$ is nonzero, and the Poincare duality pairing $H^1\times H^2\to H^3\cong {\mathbb{F}}_p$ is nondegenerate. If $h:L(p)\to L(p)$ is an orientation-reversing homotopy equivalence and $h^{\ast }a=na$, naturality givesCancelling the nonzero $t(a)$ yields $n^2\equiv -1(\mathrm{mod}p)$. Thus $-1$ must be a quadratic residue modulo $p$.

For variables $x_1,\dots ,x_m$, the elementary symmetric polynomial and power-sum symmetric polynomial arePut $E(t)={\prod }_i(1+x_{i}t)={\sum }_{r\ge 0}{e}_r{t}^r$. Its logarithmic derivative isComparing the coefficient of $t^{n-1}$ in $E^{\prime }(t)=E(t)E^{\prime }(t)/E(t)$ gives the Newton identities

The given Bockstein homomorphism is the first Steenrod square. A real line bundle is pulled back from the universal line bundle over ${\mathbb{RP}}^{\infty }$, where the degree-one generator $x$ satisfies $\beta (x)=x^2$. By naturality,

Apply the splitting principle for real vector bundles, writing the pulled-back bundle as $L_1\oplus \cdots \oplus L_r$ and $x_i=w_1(L_i)$. The pullback in cohomology is injective, $w_j(E)=e_j(x_1,\dots ,x_r)$, and the Bockstein derivation rule givesIn $e_1{e}_j$, the terms for which the index from $e_1$ already lies in the $j$-element subset give this sum, while each square-free monomial of degree $j+1$ occurs $j+1$ times. Over ${\mathbb{F}}_2$ this yields

The short exact coefficient sequenceinduces the integral Bockstein homomorphismIf $\rho :H^{i+1}(X;\mathbb{Z})\to H^{i+1}(X;\mathbb{Z}/m)$ is reduction modulo $m$, thenEquivalently, if an integral cochain $a$ lifts a modulo-$m$ cocycle and $\delta a=mb$, then $\tilde{\beta }[a]=[b]$ and $\beta [a]=[b\mathrm{mod}m]$.

For integral lifts $a,c$ of classes $x,y$, the cup product coboundary formula isDividing by $m$ and reducing modulo $m$ proves the Bockstein derivation rule

Now assume that the stated closed five-manifold $M$ exists. Its top integral cohomology makes it connected and orientable. The long exact sequence from a coefficient sequence for multiplication by $p$ shows thatand that reduction $H^3(M;\mathbb{Z})\to H^3(M;\mathbb{Z}/p)$ is an isomorphism. HenceThe same coefficient sequence gives $H^4(M;\mathbb{Z}/p)=0$.

Choose $0\ne x\in H^2(M;\mathbb{Z}/p)$ and put $y=\beta x\ne 0$. By Poincare duality over $\mathbb{Z}/p$, the pairingis nondegenerate. Both factors are one-dimensional, so $x⌣y\ne 0$. But $x⌣x=0$ because $H^4(M;\mathbb{Z}/p)=0$, while the derivation rule and graded commutativity of the cup product giveThis contradicts $p>2$. Therefore no such manifold exists.

The space $K(\mathbb{Z}/n,1)$ is a classifying space $B(\mathbb{Z}/n)$. The periodic resolution of a finite cyclic group givesThus positive odd cohomology vanishes and every positive even group is $\mathbb{Z}/n$.

Let $p$ be odd. If $p\nmid n$, transfer makes multiplication by $n$ both zero and invertible on positive-degree cohomology, soIf $p\mid n$, restriction to the cyclic Sylow $p$-subgroup and transfer giveFor $n=p$, one may take $v=\beta u$, where $\beta$ is the mod-$p$ Bockstein homomorphism.

For the second part put $A=\mathbb{Z}/p$ and $X=K(A,1)\times K(A,1)$. The long exact homotopy sequence of the homotopy fibre of $f:X\to K(A,2)$ givesand ${\pi }_k(F)=0$ for $k>1$. Therefore $F$ is a $K(G,1)$ and $|G|=p^3$.

Regard it as the fibrationWriteand write $w,t$ for the degree-one and degree-two generators of the fibre. The fibration is classified by $u_1{u}_2$, so in its cohomological Serre spectral sequenceThe Kudo transgression theorem and the mod-$p$ Bockstein giveConsequently $H^1(F;{\mathbb{F}}_p)$ has basis $u_1,u_2$, while $H^2(F;{\mathbb{F}}_p)$ has the four surviving classes represented byIf $G$ were abelian, an abelian group of order $p^3$ mapping onto $A^2$ would be either $A^3$ or $\mathbb{Z}/p^2\times A$. The first has three-dimensional $H^1(-;{\mathbb{F}}_p)$; the second has two-dimensional $H^1$ but three-dimensional $H^2$. Both contradict the dimensions just calculated. Hence $G$ is nonabelian.